Convert the binary number 1100011 (base 2) to octal (base 8). Show the 3-bit grouping from the right and the resulting octal digits.

Introduction / Context:
Binary-to-octal conversion is efficient because each octal digit corresponds exactly to three binary bits. Grouping the binary number into 3-bit chunks from the right yields the octal digits directly. This method is widely used for compactly representing long binary strings.

Given Data / Assumptions:

• Binary value: 1100011 (base 2).
• We convert without changing the numeric value, only the base representation.
• Leading zeros may be added on the left to complete a 3-bit group.

Concept / Approach:
Starting at the least significant end, partition the binary string into 3-bit groups and map each group to its octal digit: 000→0, 001→1, …, 111→7. If the leftmost group has fewer than 3 bits, pad with leading zeros to the left. Concatenate the mapped octal digits in order to form the final octal number.

Step-by-Step Solution:
Write the binary: 1 100 011 (grouped from right).Pad the leftmost group to 3 bits: 001 100 011.Map groups: 001→1, 100→4, 011→3.Concatenate octal digits: 1 4 3 → 143 (base 8).

Verification / Alternative check:
Decimal cross-check: 1100011 (base 2) = 64 + 32 + 2 + 1 = 99. Now 143 (base 8) = 164 + 48 + 3 = 64 + 32 + 3 = 99. The results match, confirming correctness.

Why Other Options Are Wrong:
140 (base 8) equals 96 decimal; 147 (base 8) equals 103 decimal; 149 is not a valid octal number because 9 is not an octal digit; “None” is incorrect because 143 is correct.

Common Pitfalls:
Forgetting to pad the leftmost group; grouping from the left instead of the right; accidentally converting to hexadecimal mapping (4-bit groups) instead of octal (3-bit groups).

Final Answer:
143